By using above basic addition/ subtraction formulae, prove that
(i) `tan (A+B)=(tan A+tan B)/(1-tan A tan B)` , (ii) `tan (A-B)=(tan A-tanB)/(1+tan A tan B)`
(iii) `sin2theta=2sintheta costheta` , (iv) `cos2theta=cos^(2)theta=1-2sin^(2)theta=2cos^(2)theta-1`
(v) `tan 2theta=(2 tan theta)/(1-tan^(2) theta)`

Explanation

(i) `tan (A+B)=(sin (A+B))/(cos (A+B))=(sin A cos B+cos A sin B)/(cos A cos B-sin A sin B)`
`=(cos A cos B((sin A)/(cosA)+(sin B)/(cos B)))/(cos A cos B(1-(sin A sin B)/(cos a cos B)))=(tan A +tan B)/(1-tan A tan B)`
(ii) `tan (A-B)=(sin (A-B))/(cos(A-B))=(sin A cos B- cos A sin B)/(cos A cos B+ sin A sin B)`
`=(cos A cos B[(sinA)/(cos A)-(sin B)/(cos B)])/(cos A cos B[1+(sin A sin B)/(cos A cos B)])=(tan A-tan B)/(1+tan A tan B)`
(iii) `sin 2theta=sin (theta+theta)=sin theta+costheta+costheta sin theta=2 sin theta costheta`
(iv) `cos2theta=cos(theta+theta)=costheta costheta-sintheta sintheta=cos^(2)theta-sin^(2)theta=1-sin^(2)theta-sin^(2)theta=1-2sin^(2)theta`
`=1-2(1-cos^(2)theta)=2 cos^(2)theta-1`
(v) `tan2theta=tan(theta+theta)=(tan theta+tan theta)/(1-tan theta tan theta)=(2 tan theta)/(1-tan^(2) theta)`